Mixing
Can two different multiplicative systems give same localisation?
$begingroup$
Can we have two different multiplicative systems $S_1$ and $S_2$ in $mathbb{Z}$ having same localisations $mathbb{Z}_{S_1}=mathbb{Z}_{S_2}$?
I have a trivial solution by taking two different subrings and adjoining $1$ to them gives required $S_1,S_2$ but that collapses the localizations to trivial rings. I wanted something more interesting. Considering multiplicative systems without $0$...
abstract-algebra ring-theory commutative-algebra localization
asked Jan 17 at 17:30
Arpan DasArpan Das
937
$endgroup$
add a comment |
$begingroup$
Can we have two different multiplicative systems $S_1$ and $S_2$ in $mathbb{Z}$ having same localisations $mathbb{Z}_{S_1}=mathbb{Z}_{S_2}$?
I have a trivial solution by taking two different subrings and adjoining $1$ to them gives required $S_1,S_2$ but that collapses the localizations to trivial rings. I wanted something more interesting. Considering multiplicative systems without $0$...
abstract-algebra ring-theory commutative-algebra localization
asked Jan 17 at 17:30
Arpan DasArpan Das
937
$endgroup$
$begingroup$
${4^n mid n in mathbb{N}}$ and ${2^n mid n in mathbb{N}}$ should yield the same localization, right? More generally, the localization at an element $p_1^{n_1} cdot ldots cdot p_r^{n_r}$ is the same as the localization of $p_1 cdot ldots cdot p_r$.
$endgroup$
– Paul K
Jan 17 at 17:36
$begingroup$
yup...$m/{4^n}=m(1/{2^n}cdot 1/{2^n})$ and $k/{2^n}={k2^n}/{4^n}$...got it
$endgroup$
– Arpan Das
Jan 17 at 17:52
add a comment |
$begingroup$
Can we have two different multiplicative systems $S_1$ and $S_2$ in $mathbb{Z}$ having same localisations $mathbb{Z}_{S_1}=mathbb{Z}_{S_2}$?
I have a trivial solution by taking two different subrings and adjoining $1$ to them gives required $S_1,S_2$ but that collapses the localizations to trivial rings. I wanted something more interesting. Considering multiplicative systems without $0$...
abstract-algebra ring-theory commutative-algebra localization
asked Jan 17 at 17:30
Arpan DasArpan Das
937
$endgroup$
Can we have two different multiplicative systems $S_1$ and $S_2$ in $mathbb{Z}$ having same localisations $mathbb{Z}_{S_1}=mathbb{Z}_{S_2}$?
I have a trivial solution by taking two different subrings and adjoining $1$ to them gives required $S_1,S_2$ but that collapses the localizations to trivial rings. I wanted something more interesting. Considering multiplicative systems without $0$...
abstract-algebra ring-theory commutative-algebra localization
abstract-algebra ring-theory commutative-algebra localization
asked Jan 17 at 17:30
Arpan DasArpan Das
937
asked Jan 17 at 17:30
Arpan DasArpan Das
937
asked Jan 17 at 17:30
Arpan DasArpan Das
937
asked Jan 17 at 17:30
Arpan DasArpan Das
937
asked Jan 17 at 17:30
Arpan DasArpan Das
937
937
$begingroup$
${4^n mid n in mathbb{N}}$ and ${2^n mid n in mathbb{N}}$ should yield the same localization, right? More generally, the localization at an element $p_1^{n_1} cdot ldots cdot p_r^{n_r}$ is the same as the localization of $p_1 cdot ldots cdot p_r$.
$endgroup$
– Paul K
Jan 17 at 17:36
$begingroup$
yup...$m/{4^n}=m(1/{2^n}cdot 1/{2^n})$ and $k/{2^n}={k2^n}/{4^n}$...got it
$endgroup$
– Arpan Das
Jan 17 at 17:52
add a comment |
$begingroup$
${4^n mid n in mathbb{N}}$ and ${2^n mid n in mathbb{N}}$ should yield the same localization, right? More generally, the localization at an element $p_1^{n_1} cdot ldots cdot p_r^{n_r}$ is the same as the localization of $p_1 cdot ldots cdot p_r$.
$endgroup$
– Paul K
Jan 17 at 17:36
$begingroup$
yup...$m/{4^n}=m(1/{2^n}cdot 1/{2^n})$ and $k/{2^n}={k2^n}/{4^n}$...got it
$endgroup$
– Arpan Das
Jan 17 at 17:52
$begingroup$
${4^n mid n in mathbb{N}}$ and ${2^n mid n in mathbb{N}}$ should yield the same localization, right? More generally, the localization at an element $p_1^{n_1} cdot ldots cdot p_r^{n_r}$ is the same as the localization of $p_1 cdot ldots cdot p_r$.
$endgroup$
– Paul K
Jan 17 at 17:36
$begingroup$
${4^n mid n in mathbb{N}}$ and ${2^n mid n in mathbb{N}}$ should yield the same localization, right? More generally, the localization at an element $p_1^{n_1} cdot ldots cdot p_r^{n_r}$ is the same as the localization of $p_1 cdot ldots cdot p_r$.
$endgroup$
– Paul K
Jan 17 at 17:36
$begingroup$
yup...$m/{4^n}=m(1/{2^n}cdot 1/{2^n})$ and $k/{2^n}={k2^n}/{4^n}$...got it
$endgroup$
– Arpan Das
Jan 17 at 17:52
$begingroup$
yup...$m/{4^n}=m(1/{2^n}cdot 1/{2^n})$ and $k/{2^n}={k2^n}/{4^n}$...got it
$endgroup$
– Arpan Das
Jan 17 at 17:52
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Here is a very general criterion. Let $A$ be a commutative ring and let $Ssubseteq A$ be a multiplicatively closed subset. Define the saturated closure of $S$ to be $$operatorname{sat}(S)={ain A: atext{ divides some element of }S}.$$ Note that $operatorname{sat}(operatorname{sat}(S))=operatorname{sat}(S)$ so this really is a "closure" operation.
Theorem: Let $S,Tsubseteq A$ be multiplicatively closed subsets of $A$. Then the localizations $A_S$ and $A_T$ are isomorphic as $A$-algebras iff $operatorname{sat}(S)=operatorname{sat}(T)$.
This gives lots of examples. For instance, in the case $A=mathbb{Z}$, you can take any nonzero integer $n$ and let $Ssubsetmathbb{Z}$ be the multiplicatively closed subset subset generated by $n$. Then $operatorname{sat}(S)$ will be the multiplicatively closed subset generated by the prime factors of $n$. So if $m$ is any other integer with the same set of prime factors as $n$ and $T$ is the multiplicatively closed subset generated by $m$, then $mathbb{Z}_Scongmathbb{Z}_T$.
Proof of Theorem: Note that in any commutative ring, any divisor of a unit is a unit. It follows that for any $A$-algebra $B$, every element of $S$ is a unit in $B$ iff every element of $operatorname{sat}(S)$ is a unit in $B$. Thus the localizations $A_S$ and $A_{operatorname{sat}(S)}$ have the same universal property as $A$-algebras, and so they are isomorphic. It follows immediately that if $operatorname{sat}(S)=operatorname{sat}(T)$ then $A_Scong A_T$.
Conversely, I claim that $operatorname{sat}(S)$ is exactly the set of elements of $A$ which are units in $A_S$. By the discussion above, every element of $operatorname{sat}(S)$ is a unit in $A_S$. Conversely, suppose $ain A$ is a unit in $A_S$. This means there exists $bin B$ and $sin S$ such that $frac{a}{1}cdotfrac{b}{s}=frac{1}{1}$ in $A_S$, which means there exists $tin S$ such that $t(ab-s)=0$. But then $abt=st$ and so $a$ divides the element $stin S$, so $ainoperatorname{sat}(S)$.
It follows that if $A_Scong A_T$ then $operatorname{sat}(S)=operatorname{sat}(T)$, since the elements of $A$ that are units in $A_S$ and $A_T$ are the same.
answered Jan 17 at 17:53
Eric WofseyEric Wofsey
187k14215344
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3077262%2fcan-two-different-multiplicative-systems-give-same-localisation%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here is a very general criterion. Let $A$ be a commutative ring and let $Ssubseteq A$ be a multiplicatively closed subset. Define the saturated closure of $S$ to be $$operatorname{sat}(S)={ain A: atext{ divides some element of }S}.$$ Note that $operatorname{sat}(operatorname{sat}(S))=operatorname{sat}(S)$ so this really is a "closure" operation.
Theorem: Let $S,Tsubseteq A$ be multiplicatively closed subsets of $A$. Then the localizations $A_S$ and $A_T$ are isomorphic as $A$-algebras iff $operatorname{sat}(S)=operatorname{sat}(T)$.
This gives lots of examples. For instance, in the case $A=mathbb{Z}$, you can take any nonzero integer $n$ and let $Ssubsetmathbb{Z}$ be the multiplicatively closed subset subset generated by $n$. Then $operatorname{sat}(S)$ will be the multiplicatively closed subset generated by the prime factors of $n$. So if $m$ is any other integer with the same set of prime factors as $n$ and $T$ is the multiplicatively closed subset generated by $m$, then $mathbb{Z}_Scongmathbb{Z}_T$.
Proof of Theorem: Note that in any commutative ring, any divisor of a unit is a unit. It follows that for any $A$-algebra $B$, every element of $S$ is a unit in $B$ iff every element of $operatorname{sat}(S)$ is a unit in $B$. Thus the localizations $A_S$ and $A_{operatorname{sat}(S)}$ have the same universal property as $A$-algebras, and so they are isomorphic. It follows immediately that if $operatorname{sat}(S)=operatorname{sat}(T)$ then $A_Scong A_T$.
Conversely, I claim that $operatorname{sat}(S)$ is exactly the set of elements of $A$ which are units in $A_S$. By the discussion above, every element of $operatorname{sat}(S)$ is a unit in $A_S$. Conversely, suppose $ain A$ is a unit in $A_S$. This means there exists $bin B$ and $sin S$ such that $frac{a}{1}cdotfrac{b}{s}=frac{1}{1}$ in $A_S$, which means there exists $tin S$ such that $t(ab-s)=0$. But then $abt=st$ and so $a$ divides the element $stin S$, so $ainoperatorname{sat}(S)$.
It follows that if $A_Scong A_T$ then $operatorname{sat}(S)=operatorname{sat}(T)$, since the elements of $A$ that are units in $A_S$ and $A_T$ are the same.
answered Jan 17 at 17:53
Eric WofseyEric Wofsey
187k14215344
$endgroup$
add a comment |
$begingroup$
Here is a very general criterion. Let $A$ be a commutative ring and let $Ssubseteq A$ be a multiplicatively closed subset. Define the saturated closure of $S$ to be $$operatorname{sat}(S)={ain A: atext{ divides some element of }S}.$$ Note that $operatorname{sat}(operatorname{sat}(S))=operatorname{sat}(S)$ so this really is a "closure" operation.
Theorem: Let $S,Tsubseteq A$ be multiplicatively closed subsets of $A$. Then the localizations $A_S$ and $A_T$ are isomorphic as $A$-algebras iff $operatorname{sat}(S)=operatorname{sat}(T)$.
This gives lots of examples. For instance, in the case $A=mathbb{Z}$, you can take any nonzero integer $n$ and let $Ssubsetmathbb{Z}$ be the multiplicatively closed subset subset generated by $n$. Then $operatorname{sat}(S)$ will be the multiplicatively closed subset generated by the prime factors of $n$. So if $m$ is any other integer with the same set of prime factors as $n$ and $T$ is the multiplicatively closed subset generated by $m$, then $mathbb{Z}_Scongmathbb{Z}_T$.
Proof of Theorem: Note that in any commutative ring, any divisor of a unit is a unit. It follows that for any $A$-algebra $B$, every element of $S$ is a unit in $B$ iff every element of $operatorname{sat}(S)$ is a unit in $B$. Thus the localizations $A_S$ and $A_{operatorname{sat}(S)}$ have the same universal property as $A$-algebras, and so they are isomorphic. It follows immediately that if $operatorname{sat}(S)=operatorname{sat}(T)$ then $A_Scong A_T$.
Conversely, I claim that $operatorname{sat}(S)$ is exactly the set of elements of $A$ which are units in $A_S$. By the discussion above, every element of $operatorname{sat}(S)$ is a unit in $A_S$. Conversely, suppose $ain A$ is a unit in $A_S$. This means there exists $bin B$ and $sin S$ such that $frac{a}{1}cdotfrac{b}{s}=frac{1}{1}$ in $A_S$, which means there exists $tin S$ such that $t(ab-s)=0$. But then $abt=st$ and so $a$ divides the element $stin S$, so $ainoperatorname{sat}(S)$.
It follows that if $A_Scong A_T$ then $operatorname{sat}(S)=operatorname{sat}(T)$, since the elements of $A$ that are units in $A_S$ and $A_T$ are the same.
answered Jan 17 at 17:53
Eric WofseyEric Wofsey
187k14215344
$endgroup$
add a comment |
$begingroup$
Here is a very general criterion. Let $A$ be a commutative ring and let $Ssubseteq A$ be a multiplicatively closed subset. Define the saturated closure of $S$ to be $$operatorname{sat}(S)={ain A: atext{ divides some element of }S}.$$ Note that $operatorname{sat}(operatorname{sat}(S))=operatorname{sat}(S)$ so this really is a "closure" operation.
Theorem: Let $S,Tsubseteq A$ be multiplicatively closed subsets of $A$. Then the localizations $A_S$ and $A_T$ are isomorphic as $A$-algebras iff $operatorname{sat}(S)=operatorname{sat}(T)$.
This gives lots of examples. For instance, in the case $A=mathbb{Z}$, you can take any nonzero integer $n$ and let $Ssubsetmathbb{Z}$ be the multiplicatively closed subset subset generated by $n$. Then $operatorname{sat}(S)$ will be the multiplicatively closed subset generated by the prime factors of $n$. So if $m$ is any other integer with the same set of prime factors as $n$ and $T$ is the multiplicatively closed subset generated by $m$, then $mathbb{Z}_Scongmathbb{Z}_T$.
Proof of Theorem: Note that in any commutative ring, any divisor of a unit is a unit. It follows that for any $A$-algebra $B$, every element of $S$ is a unit in $B$ iff every element of $operatorname{sat}(S)$ is a unit in $B$. Thus the localizations $A_S$ and $A_{operatorname{sat}(S)}$ have the same universal property as $A$-algebras, and so they are isomorphic. It follows immediately that if $operatorname{sat}(S)=operatorname{sat}(T)$ then $A_Scong A_T$.
Conversely, I claim that $operatorname{sat}(S)$ is exactly the set of elements of $A$ which are units in $A_S$. By the discussion above, every element of $operatorname{sat}(S)$ is a unit in $A_S$. Conversely, suppose $ain A$ is a unit in $A_S$. This means there exists $bin B$ and $sin S$ such that $frac{a}{1}cdotfrac{b}{s}=frac{1}{1}$ in $A_S$, which means there exists $tin S$ such that $t(ab-s)=0$. But then $abt=st$ and so $a$ divides the element $stin S$, so $ainoperatorname{sat}(S)$.
It follows that if $A_Scong A_T$ then $operatorname{sat}(S)=operatorname{sat}(T)$, since the elements of $A$ that are units in $A_S$ and $A_T$ are the same.
answered Jan 17 at 17:53
Eric WofseyEric Wofsey
187k14215344
$endgroup$
Here is a very general criterion. Let $A$ be a commutative ring and let $Ssubseteq A$ be a multiplicatively closed subset. Define the saturated closure of $S$ to be $$operatorname{sat}(S)={ain A: atext{ divides some element of }S}.$$ Note that $operatorname{sat}(operatorname{sat}(S))=operatorname{sat}(S)$ so this really is a "closure" operation.
Theorem: Let $S,Tsubseteq A$ be multiplicatively closed subsets of $A$. Then the localizations $A_S$ and $A_T$ are isomorphic as $A$-algebras iff $operatorname{sat}(S)=operatorname{sat}(T)$.
This gives lots of examples. For instance, in the case $A=mathbb{Z}$, you can take any nonzero integer $n$ and let $Ssubsetmathbb{Z}$ be the multiplicatively closed subset subset generated by $n$. Then $operatorname{sat}(S)$ will be the multiplicatively closed subset generated by the prime factors of $n$. So if $m$ is any other integer with the same set of prime factors as $n$ and $T$ is the multiplicatively closed subset generated by $m$, then $mathbb{Z}_Scongmathbb{Z}_T$.
Proof of Theorem: Note that in any commutative ring, any divisor of a unit is a unit. It follows that for any $A$-algebra $B$, every element of $S$ is a unit in $B$ iff every element of $operatorname{sat}(S)$ is a unit in $B$. Thus the localizations $A_S$ and $A_{operatorname{sat}(S)}$ have the same universal property as $A$-algebras, and so they are isomorphic. It follows immediately that if $operatorname{sat}(S)=operatorname{sat}(T)$ then $A_Scong A_T$.
Conversely, I claim that $operatorname{sat}(S)$ is exactly the set of elements of $A$ which are units in $A_S$. By the discussion above, every element of $operatorname{sat}(S)$ is a unit in $A_S$. Conversely, suppose $ain A$ is a unit in $A_S$. This means there exists $bin B$ and $sin S$ such that $frac{a}{1}cdotfrac{b}{s}=frac{1}{1}$ in $A_S$, which means there exists $tin S$ such that $t(ab-s)=0$. But then $abt=st$ and so $a$ divides the element $stin S$, so $ainoperatorname{sat}(S)$.
It follows that if $A_Scong A_T$ then $operatorname{sat}(S)=operatorname{sat}(T)$, since the elements of $A$ that are units in $A_S$ and $A_T$ are the same.
answered Jan 17 at 17:53
Eric WofseyEric Wofsey
187k14215344
answered Jan 17 at 17:53
Eric WofseyEric Wofsey
187k14215344
answered Jan 17 at 17:53
Eric WofseyEric Wofsey
187k14215344
answered Jan 17 at 17:53
Eric WofseyEric Wofsey
187k14215344
187k14215344
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3077262%2fcan-two-different-multiplicative-systems-give-same-localisation%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
${4^n mid n in mathbb{N}}$ and ${2^n mid n in mathbb{N}}$ should yield the same localization, right? More generally, the localization at an element $p_1^{n_1} cdot ldots cdot p_r^{n_r}$ is the same as the localization of $p_1 cdot ldots cdot p_r$.
$endgroup$
– Paul K
Jan 17 at 17:36
$begingroup$
yup...$m/{4^n}=m(1/{2^n}cdot 1/{2^n})$ and $k/{2^n}={k2^n}/{4^n}$...got it
$endgroup$
– Arpan Das
Jan 17 at 17:52